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mOOC
qMRLab

Appendix A

Double Angle technique

NeuroPoly Lab, Polytechnique Montreal, Quebec, Canada
ei2α=ei(α+α)=eiα+iα=eiαeiα=(cos(α)+isin(α))(cos(α)+isin(α))=cos(α)cos(α)+icos(α)sin(α)+isin(α)cos(α)+i2sin(α)sin(α)=cos(α)cos(α)+icos(α)sin(α)+isin(α)cos(α)+(−1)sin(α)sin(α)=cos(α)cos(α)+icos(α)sin(α)+isin(α)cos(α)−sin(α)sin(α)=cos2(α)+icos(α)sin(α)+isin(α)cos(α)−sin2(α)=(cos2(α)−sin2(α))+i(cos(α)sin(α)+sin(α)cos(α))=(cos2(α)−sin2(α))+i(sin(α)cos(α)+sin(α)cos(α))ei2α=(cos2(α)−sin2(α))+i(2sin(α)cos(α))cos(2α)+isin(2α)=(cos2(α)−sin2(α))+i(2sin(α)cos(α))For z∈C and q∈C,if z=qthen Re(z)=Re(q) and Im(z)=Im(q)thus,Im(cos(2α)+isin(2α))=Im((cos2(α)−sin2(α))+i(2sin(α)cos(α)))sin(2α)=2sin(α)cos(α)Q.E.D.\begin{split} \text{e}^{i2\alpha} &= \text{e}^{i\left( \alpha+\alpha \right)} \\ &= \text{e}^{i\alpha+i\alpha} \\ &= \text{e}^{i\alpha}\text{e}^{i\alpha} \\ &= \left( \text{cos}\left( \alpha \right)+i\text{sin}\left( \alpha \right) \right)\left( \text{cos}\left( \alpha \right)+i\text{sin}\left( \alpha \right) \right) \\ &= \text{cos}\left( \alpha \right)\text{cos}\left( \alpha \right)+i\text{cos}\left( \alpha \right)\text{sin}\left( \alpha \right)+i\text{sin}\left( \alpha \right)\text{cos}\left( \alpha \right)+i^{2}\text{sin}\left( \alpha \right)\text{sin}\left( \alpha \right)\\ &= \text{cos}\left( \alpha \right)\text{cos}\left( \alpha \right)+i\text{cos}\left( \alpha \right)\text{sin}\left( \alpha \right)+i\text{sin}\left( \alpha \right)\text{cos}\left( \alpha \right)+\left( -1 \right)\text{sin}\left( \alpha \right)\text{sin}\left( \alpha \right)\\ &= \text{cos}\left( \alpha \right)\text{cos}\left( \alpha \right)+i\text{cos}\left( \alpha \right)\text{sin}\left( \alpha \right)+i\text{sin}\left( \alpha \right)\text{cos}\left( \alpha \right)-\text{sin}\left( \alpha \right)\text{sin}\left( \alpha \right)\\ &= \text{cos}^{2}\left( \alpha \right)+i\text{cos}\left( \alpha \right)\text{sin}\left( \alpha \right)+i\text{sin}\left( \alpha \right)\text{cos}\left( \alpha \right)-\text{sin}^{2}\left( \alpha \right)\\ &= \left( \text{cos}^{2}\left( \alpha \right)-\text{sin}^{2}\left( \alpha \right)\right)+i\left( \text{cos}\left( \alpha \right)\text{sin}\left( \alpha \right) +\text{sin}\left( \alpha \right)\text{cos}\left( \alpha \right) \right)\\ &= \left( \text{cos}^{2}\left( \alpha \right)-\text{sin}^{2}\left( \alpha \right)\right)+i\left( \text{sin}\left( \alpha \right) \text{cos}\left( \alpha \right)+\text{sin}\left( \alpha \right)\text{cos}\left( \alpha \right) \right)\\ \text{e}^{i2\alpha} &= \left( \text{cos}^{2}\left( \alpha \right)-\text{sin}^{2}\left( \alpha \right)\right)+i\left( 2\text{sin}\left( \alpha \right) \text{cos}\left( \alpha \right) \right)\\ \text{cos}\left( 2\alpha \right)+i\text{sin}\left( 2\alpha \right) &= \left( \text{cos}^{2}\left( \alpha \right)-\text{sin}^{2}\left( \alpha \right)\right)+i\left( 2\text{sin}\left( \alpha \right) \text{cos}\left( \alpha \right) \right)\\ \end{split} \\ \\ \text{For }z \in \mathbb{C} \text{ and }q \in \mathbb{C}\text{,}\\ \text{if }z=q \\ \text{then }\text{Re}\left( z \right)=\text{Re}\left( q \right) \\ \text{ and }\text{Im}\left( z \right)=\text{Im}\left( q \right) \\ \text{thus,} \\ \\ \begin{split} \text{Im}\left( \text{cos}\left( 2\alpha \right)+i\text{sin}\left( 2\alpha \right) \right) &= \text{Im}\left( \left( \text{cos}^{2}\left( \alpha \right)-\text{sin}^{2}\left( \alpha \right)\right)+i\left( 2\text{sin}\left( \alpha \right) \text{cos}\left( \alpha \right) \right) \right)\\ \text{sin}\left( 2\alpha \right) &= 2\text{sin}\left( \alpha \right) \text{cos}\left( \alpha \right) \\ \end{split}\\ Q.E.D.
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Double Angle Methods
Gradient echo and 180 degree spin echo method
Notebooks
Figure 4 1 1